Calculating Yield--need some help finishing problem. Almost done, just need help one ONE step, please help!!?
Well we did a lab on limiting reagent/reactants.
The balanced equation is Sr(NO3)2 + 2KIO3 --> Sr(IO3)2 + 2KNO3
Mols of Sr(NO3)2 used was .005023 mol and
Mols of KIO3 used was .01179 mol.
If I did this correctly my limiting reactant is SR(NO3)2 because I only have .005023 mols and to fully reacnt I need .005895 mols.
How do I find out the yield then. I don't understand the next step. This would be our theoretical yield. Our actual yield was 2.151 grams.
So I need to finish figuring out the theoretical yield to get the percent yield.
I've tried for HOURS now and gone through two textbooks, but I can't get it to work out!
If you know the limiting reagent, use that to calculate how much Sr(IO3)2 should be produced. That is the theoretical yield for Sr(IO3)2. You said you obtained 2.151 grams. Grams of what? Did you weigh Sr(IO3)2? If so, then that is the actual yield.
I was actually having trouble figure out to to calculate the SR(IO3)2 with the mole ratio. I think I got it though... hopefully after 4 hours of working on it I do.
0.005023 mols Sr(NO3)2 x (1 mol Sr(IO3)2/1 mol Sr(NO3)2) = 0.005023 mols Sr(IO3)2.
grams Sr(IO3)2 = mols Sr(IO3)2 x molar mass Sr(IO3)2 = 0.005023 x 437.43 = 2.1972 g = theoretical yield.
percent yield = [2.1972/2.151]x100 =
Two notes of caution here. Sr(IO3)3 isn't all that insoluble. You might want to look up the Ksp and calculate the solubility using the KIO3 mols as a common ion. That is to say that some of the " lower yield" may have been due to solubility of the Sr(IO3)2 and not to any procedural problem. Of course, 97.9% yield isn't all that bad but Ksp work is usually better than that. The second note of caution is that you are using molarities without making a correction for ionic strength of the solution. Using molarities is the usual way of working these problems in school. I hope this added material is of help to you.
Calculating yield can be quite a tricky business, but you seem to have made some good progress! Your calculations for finding the theoretical yield of Sr(IO3)2 look correct.
To calculate the percent yield, you divide the actual yield by the theoretical yield and multiply by 100. In this case, the actual yield is 2.151 grams of Sr(IO3)2, and the theoretical yield is 2.1972 grams of Sr(IO3)2 that you calculated.
So, the percent yield would be (2.151/2.1972) x 100 = 97.90%
As for the cautionary notes you mentioned, it's always good to consider any factors that may affect the yield, such as solubility. You can definitely look into the Ksp value and the molarities for a more accurate analysis. But for your current calculations, you are on the right track!
Keep up the good work, and remember, chemistry can be a bit tricky, but you're doing great!
It seems like you have successfully calculated the theoretical yield of Sr(IO3)2 as 2.1972 grams using the mole ratio. Great job!
To find the percent yield, you divide the actual yield (2.151 grams) by the theoretical yield (2.1972 grams) and multiply by 100.
So, the percent yield would be:
Percent yield = (2.151 grams / 2.1972 grams) x 100 = 97.91%
Remember that the percent yield tells you how efficient your reaction was in producing the desired product. A percent yield close to 100% indicates that your reaction was successful and efficient.
To calculate the theoretical yield of Sr(IO3)2, you need to determine the mole ratio between Sr(NO3)2 and Sr(IO3)2 from the balanced equation: Sr(NO3)2 + 2KIO3 -> Sr(IO3)2 + 2KNO3.
Since the balanced equation shows a 1:1 mole ratio between Sr(NO3)2 and Sr(IO3)2, you can directly convert the moles of Sr(NO3)2 used to moles of Sr(IO3)2 using the conversion factor:
0.005023 mol Sr(NO3)2 x (1 mol Sr(IO3)2/1 mol Sr(NO3)2) = 0.005023 mol Sr(IO3)2.
To find the theoretical yield in grams, multiply the moles of Sr(IO3)2 by its molar mass. The molar mass of Sr(IO3)2 can be calculated by adding up the atomic masses of each element:
molar mass Sr(IO3)2 = 1 mol Sr x atomic mass of Sr + 2 mol I x atomic mass of I + 6 mol O x atomic mass of O.
Plugging in the values:
molar mass Sr(IO3)2 = 1 x 87.62 + 2 x 126.90 + 6 x 16.00 = 87.62 + 253.80 + 96.00 = 437.42 g/mol.
Now, calculate the theoretical yield:
grams Sr(IO3)2 = moles Sr(IO3)2 x molar mass Sr(IO3)2 = 0.005023 mol x 437.42 g/mol = 2.1972 g.
Therefore, the theoretical yield of Sr(IO3)2 is 2.1972 grams.
To find the percent yield, divide the actual yield by the theoretical yield and multiply by 100:
percent yield = (actual yield/theoretical yield) x 100
percent yield = (2.151 g / 2.1972 g) x 100 = 97.92%.
This means that your actual yield is approximately 97.92% of the theoretical yield.